A steel scale is to be prepared such that the millimeter intervals are to be accurate within `6 xx10^(-5)mm`. The maximum temperature variation form the temperature of calibration during the reading of the millimeter marks is `(alpha = 12 xx 10^(-6)//"^(@)C)`

Options

A. `4.0"^(@)C`
B. `4.5 "^(@)C`
C. `5.0"^(@)C`
D. `5.5 "^(@)C`

Answer

To solve the problem step by step, we will use the formula for linear expansion and the given data.

### Step 1: Identify the given values
- Length of the interval (L) = 1 mm = \(1 \times 10^{-3}\) m (convert to meters for consistency)
- Change in length (ΔL) = \(6 \times 10^{-5}\) mm = \(6 \times 10^{-8}\) m (convert to meters)
- Coefficient of linear expansion (α) = \(12 \times 10^{-6} \, \text{°C}^{-1}\)

### Step 2: Write down the formula for linear expansion
The formula for linear expansion is given by:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
where:
- ΔL = change in length
- L = original length
- α = coefficient of linear expansion
- ΔT = change in temperature

### Step 3: Rearrange the formula to solve for ΔT
We need to isolate ΔT:
\[ \Delta T = \frac{\Delta L}{L \cdot \alpha} \]

### Step 4: Substitute the known values into the equation
Substituting the values we have:
\[ \Delta T = \frac{6 \times 10^{-8} \, \text{m}}{(1 \times 10^{-3} \, \text{m}) \cdot (12 \times 10^{-6} \, \text{°C}^{-1})} \]

### Step 5: Calculate ΔT
Calculating the denominator:
\[ (1 \times 10^{-3} \, \text{m}) \cdot (12 \times 10^{-6} \, \text{°C}^{-1}) = 12 \times 10^{-9} \, \text{m°C}^{-1} \]
Now substituting back into the equation:
\[ \Delta T = \frac{6 \times 10^{-8}}{12 \times 10^{-9}} = \frac{6}{12} \times 10^{1} = 0.5 \times 10^{1} = 5 \, \text{°C} \]

### Step 6: Conclusion
The temperature variation (ΔT) is \(5 \, \text{°C}\).

### Final Answer
\[ \Delta T = 5 \, \text{°C} \]

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